Problem: Carbon- $14$ is an element which loses half of its mass every $5730$ years. The following function gives the mass, in grams, of a sample of carbon- $14$ after $t$ years: $M(t)=65\cdot e^{-0.00012t}$ What is the instantaneous rate of change of sample's mass after $1$ year? Choose 1 answer: Choose 1 answer: (Choice A) A $-0.0078$ years per gram (Choice B) B $-0.0078$ grams per year (Choice C) C $-0.78$ years per gram (Choice D) D $-0.78$ grams per year
Understanding the problem The function that represents the instantaneous rate of change of $M(t)$ is its derivative, $M'(t)$. Therefore, the instantaneous rate of change of the sample's mass after $1$ year is $M'(1)$. Let's find $M'(t)$ and evaluate it at $t=1$. Finding $M'(t)$ $M'(t)=-0.0078\cdot e^{-0.00012t}$ Finding $M'(1)$ $\begin{aligned} M'({1})&=-0.0078\cdot e^{-0.00012({1})} \\\\ &=-0.0078\cdot e^{-0.00012} \\\\ &\approx -0.0078 \end{aligned}$ Interpreting units $M(t)$ is the sample's mass in ${\text{grams}}$ after $t$ ${\text{years}}$. Therefore, we measure its rate of change in ${\text{grams}}$ per ${\text{year}}$. In conclusion, the instantaneous rate of change of the sample's mass after $1$ year is $-0.0078$ grams per year. The rate of change is negative because the sample is decaying.